Math 310: Hour Exam 2 (Solutions) Prof. S. Smith: Fri 10 Nov 2000

You must SHOW WORK to receive credit. Wherever you use a calculator, write “used calculator”.

Problem 1:

(a) Give the matrix A representing (in the standard basis) the linear transformation L : R^{2} → R^{2} defined by L((x_{1}, x_{2})^{T}) = (3x_{1} − x_{2}, x_{1} + 2x_{2})^{T}.

Apply L to standard basis, put into columns to get A = ( \begin{matrix}3 & −1 \\ 1 & 2 \end{matrix} )

(b) Now give the matrix B for the same L as in part (a), but using the basis (1, 1)^{T}, (1, 2)^{T}. Either compute directly with respect to this “new” basis; or use change-of-basis matrix

from “new” to “old” basis given by S = ( \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} ), and mutliply out B = S^{−1}AS:

( \begin{matrix}1 & −3 \\ 1 & 4 \end{matrix} ) = ( \begin{matrix}2 & −1 \\ −1 & 1 \end{matrix} ) ( \begin{matrix}3 & −1 \\ 1 & 2 \end{matrix} ) ( \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} )

Problem 2: (a) The formula ⟨f(x), g(x)⟩ = ∫_{−1}^{1} f(x)g(x) dx defines an inner product on the space P_{<2} of polynomials of degree less than 2. Show that the functions 1 and x are orthogonal in this inner product. Determine the length ||1|| fo the function 1 in this inner product.

⟨1, x⟩ = ∫_{−1}^{1} 1.x dx = [\frac{x^{2}}{2} ]_{−1}^{1} = \frac{1}{2} − \frac{1}{2} = 0, as desired. ⟨1, 1⟩ = ∫_{−1}^{1} 1.1 dx = [x]_{−1}^{1} = 1 − (−1) = 2, so ||1|| = \sqrt{2}.

(b) Find a basis for the orthogonal complement to the row space of A = ( \begin{matrix}1 & 2 & 1 & 2 \\ 1 & 3 & 1 & 3 \end{matrix} ).

By Thm 5.2.1 this is just the nullspace of A. Compute rref(A): A = ( \begin{matrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{matrix} ).

Then variables x_{3}, x_{4} are free, so solutions are (−x_{3}, −x_{4}, x_{3}, x_{4}). Choosing the free variables in the “standard basis” way, a basis is given by the two vectors (−1, 0, 1, 0) and (0, −1, 0, 1).

Problem 3:

(a) Find all solutions of the system Ax = b given by:

( \begin{matrix}1 & 2 \\ 2 & 4 \\ −1 & −2 \end{matrix} ) ( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) = ( \begin{matrix}3 \\ 2 \\ 1 \end{matrix} ) .

The row-reduced echelon form of the augmented matrix [A|b] is: ( \begin{matrix}1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} ).

The second row says 0 = 1, so there are no solutions.

(b) Find the projection p of b into the column space of A.

This is “least squares”: Multiply A^{T} by the augmented matrix [A|b] to get normal equations

( \begin{matrix}1 & 2 & −1 \\ 2 & 4 & −2 \end{matrix} ) ( \begin{matrix}1 & 2 & 3 \\ 2 & 4 & 2 \\ −1 & −2 & 1 \end{matrix} ) = ( \begin{matrix}6 & 12 & 6 \\ 12 & 24 & 12 \end{matrix} ). Compute rref: ( \begin{matrix}1 & 2 & 1 \\ 0 & 0 & 0 \end{matrix} ).

There are an infinite number of solutions for x̂, of form (1 − 2x_{2}, x_{2})^{T}; easiest to take x_{2} = 0, so x̂ = (1, 0)^{T}. Then p = Ax̂ = (1, 2, −1)^{T}.

Problem 4:

(a) Let S be the subspace of R^{3} (let’s use row vectors for convenience) spanned by v_{1} = (1, 1, 0) and v_{2} = (0, 1, 1). Use the Gram-Schmidt process to find an orthonormal basis for S.

First get orthogonal: use q_{1} = v_{1} = (1, 1, 0) and then

q_{2} = v_{2} − [(v_{2} · q_{1})/(q_{1} · q_{1})]q_{1} = (0, 1, 1) − [1/2](1, 1, 0) = (−1/2, 1/2, 1).

To make orthoNORMAL, divide by lengths to get u_{1} = \frac{1}{\sqrt{2}}(1, 1, 0) and u_{2} = \frac{1}{\sqrt{6}}(−1, 1, 2).

(b) Now give the projection p of b = (1, 1, 1) in the space S of part (a).

Could do by least squares; or, as in Thm 5.5.7:

p = (u_{1} · b)u_{1} + (u_{2} · b)u_{2} = \frac{2}{\sqrt{2}} \frac{1}{\sqrt{2}}(1, 1, 0) + \frac{2}{\sqrt{6}} \frac{1}{\sqrt{6}}(−1, 1, 2) = (1, 1, 0) + \frac{1}{3}(−1, 1, 2) = \frac{2}{3}(1, 2, 1).

Problem 5:

(a) The “trace” of a 2 × 2 matrix A is the sum of the elements on its main diagonal, namely T r(A) = A_{1,1} + A_{2,2}. Show that the trace function T r is a linear transformation on R^{2×2} (the image space is just scalars R).

(+) T r(A + B) = (A + B)_{1,1} + (A + B)_{2,2} = (A_{1,1} + B_{1,1}) + (A_{2,2} + B_{2,2})

= (A_{1,1} + A_{2,2}) + (B_{1,1} + B_{2,2}) = T r(A) + T r(B). (Or, you can write out general 2 × 2 matrices in full).

(sc.mult.) T r(cA) = (cA)_{1,1} + (cA)_{2,2} = cA_{1,1} + cA_{2,2} = c(A_{1,1} + A_{2,2}) = cT r(A).

(b) Suppose A is a 3 × 5 matrix, and L is the usual linear transformation from R^{5} to R^{3} given by left multiplication: L(x) = Ax. If A has rank 3, what is the dimension of the kernel of L ? Why ?

The kernel of L is just the nullspace N(A) of A, so its dimension is the nullity of A.

Then (by Thm 3.6.4) #(cols ofA) = 5 = rank(A) + nullity(A), so nullity(A) = 2.